Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using vi = vector<long long int>;
using vvi = vector<vi>;
using vb = vector<bool>;
using vc = vector<char>;
using vs = vector<string>;
using vld = vector<long double>;
using pii = pair<long long int, long long int>;
using psi = pair<string, long long int>;
using pci = pair<char, long long int>;
using vpii = vector<pii>;
long long int const mod = 1e9 + 7;
long long int const maxn = 1e5 + 5;
long long int const inf = 1e18;
long long int add(long long int a, long long int b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long int mul(long long int a, long long int b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long int powm(long long int x, long long int n, long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long int modinverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
bool prime(long long int x) {
  if (x <= 1) return false;
  for (int i = 2; i <= sqrt(x); i++)
    if (x % i == 0) return false;
  return true;
}
long long int divisor(long long int x) {
  long long int cnt = 0;
  for (int i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      if (i != x / i)
        cnt += 2;
      else
        cnt += 1;
    }
  }
  return cnt;
}
vector<long long int> sieve(long long int n) {
  bool prim[n + 1];
  memset(prim, true, sizeof(prim));
  for (long long int p = 2; p * p <= n; p++) {
    if (prim[p] == true) {
      for (int i = p * p; i <= n; i += p) prim[i] = false;
    }
  }
  vector<long long int> v;
  for (int i = 2; i <= n; i++)
    if (prim[i]) v.push_back(i);
  return v;
}
int const N = 1e5 + 5;
int n, m, k, q;
int const logN = 19;
vector<int> adj[N];
vector<int> a[N];
vector<int> tin(N), tout(N), P(N);
vector<int> dp[N][logN + 1];
int up[N][logN + 1];
int timer;
void merge(vector<int> &to, vector<int> &u, vector<int> &v,
           const int limit = 10) {
  int i = 0, j = 0;
  while ((int)to.size() < limit and i < (int)u.size() and j < (int)v.size()) {
    to.push_back((u[i] < v[j]) ? u[i++] : v[j++]);
  }
  while ((int)to.size() < limit and i < (int)u.size()) to.push_back(u[i++]);
  while ((int)to.size() < limit and j < (int)v.size()) to.push_back(v[j++]);
}
void binary_lifting(int u, int par) {
  up[u][0] = par;
  tin[u] = ++timer;
  for (int i = 0; i < min(10, (int)a[par].size()); i++)
    dp[u][0].push_back(a[par][i]);
  for (int h = 1; h <= logN; h++) {
    up[u][h] = up[up[u][h - 1]][h - 1];
    merge(dp[u][h], dp[u][h - 1], dp[up[u][h - 1]][h - 1]);
  }
  for (auto v : adj[u]) {
    if (v == par) continue;
    binary_lifting(v, u);
  }
  tout[u] = ++timer;
}
bool is_ancestor(int u, int v) {
  return (tin[u] <= tin[v] and tout[u] >= tout[v] and u and v);
}
int lca(int u, int v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (int h = logN; h >= 0; h--) {
    if (!is_ancestor(up[u][h], v) and up[u][h]) u = up[u][h];
  }
  return up[u][0];
}
void all_capitals(set<int> &st, int v, int anc, int limit) {
  for (int h = logN; h >= 0; h--) {
    if (!is_ancestor(up[v][h], anc) and up[v][h]) {
      int it = 1;
      for (auto id : dp[v][h]) {
        st.insert(id);
        if (it++ >= limit) break;
      }
      while ((int)st.size() > limit) st.erase(--st.end());
      v = up[v][h];
    }
  }
}
void all_capitals(set<int> &st, int v, int limit) {
  int it = 1;
  for (auto id : a[v]) {
    st.insert(id);
    if (it++ >= limit) break;
  }
  while ((int)st.size() > limit) st.erase(--st.end());
}
void solve() {
  cin >> n >> m >> q;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int capital;
    cin >> capital;
    a[capital].push_back(i);
  }
  binary_lifting(1, 0);
  while (q--) {
    int u, v, g;
    cin >> u >> v >> g;
    if (u == v) {
      cout << min(g, (int)a[u].size()) << " ";
      for (int i = 0; i < min(g, (int)a[u].size()); i++) {
        cout << a[u][i] << " ";
      }
      cout << "\n";
      continue;
    }
    set<int> ans;
    if (!is_ancestor(u, v)) swap(u, v);
    if (!(is_ancestor(u, v) or is_ancestor(v, u))) {
      int anc = lca(u, v);
      all_capitals(ans, u, anc, g);
      all_capitals(ans, v, anc, g);
      all_capitals(ans, u, g);
      all_capitals(ans, v, g);
      all_capitals(ans, anc, g);
    } else {
      all_capitals(ans, v, u, g);
      all_capitals(ans, v, g);
      all_capitals(ans, u, g);
    }
    cout << (int)ans.size() << " ";
    for (auto id : ans) cout << id << " ";
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  clock_t zzzz = clock();
  long long int t;
  t = 1;
  while (t--) {
    solve();
  }
  fprintf(stderr, "Total Time: %.3f\n",
          (double)(clock() - zzzz) / CLOCKS_PER_SEC),
      fflush(stderr);
  return 0;
}